$\overline{AB}$ = $\sqrt{149}$ $\overline{BC} = {?}$ $A$ $C$ $B$ $\sqrt{149}$ $?$ $ \sin( \angle BAC ) = \frac{10\sqrt{149} }{149}, \cos( \angle BAC ) = \frac{7\sqrt{149} }{149}, \tan( \angle BAC ) = \dfrac{10}{7}$
$\overline{AB}$ is the hypotenuse $\overline{BC}$ is opposite to $\angle BAC$ SOH CAH TOA We know the hypotenuse and need to solve for the opposite side so we can use the sine function (SOH) $ \sin( \angle BAC ) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{\overline{BC}}{\overline{AB}}= \frac{\overline{BC}}{\sqrt{149}} $ $ \overline{BC}=\sqrt{149} \cdot \sin( \angle BAC ) = \sqrt{149} \cdot \frac{10\sqrt{149} }{149} = 10$